a body is at equlibrium under the action of three forces.one force is 10n acting due east and one is 5n in the direction 60degree north east.what is the magnitude and direction of the third force.

Answer 1

10 + 5[60o] + F = 0.
10 + 5*Cos60 + i5*sin60 + F = 0.
10 + 2.5 + 4.33i + F = 0.
12.5 + 4.33i + F = 0.
F = -12.5-4.33i = 13.2N[W19.1oS]